If the electric potential is location dependent, why do charges loose it when passed through a heavy resistor in a circuit?

Question

I really cannot understand this: I know through reading that unlike electric potential energy, which is charge dependent, the electric potential is purely location dependent. For example: If at a point, the electric potential is $5V$, then $1C$ of charge will have $5J$ of energy, and $10C$ of charge will have $50J$ of energy in a particular electric field. So, this is my question:

If electric potential is dependent upon position within the electric field, then why do we say that upon passing through a heavy resistor(like a bulb) in a circuit, the charge looses electric potential? Isn't this incorrect? Shouldn't a charge loose energy, instead of potential, because potential is lost constantly as the charge moves through the ciruit?

And one more question: When a charge moves through a bulb(or any heavy resistor), what happens to the electric field?

Suppose, a charge moves through a bulb, looses some energy, and again, due to the electric field, gains some energy back.

So, what effect this has on the battery? Why does a battery run out just because charge looses energy to light up the bulb(or any other appliance)? According to me, a battery should loose its "capability" of pumping charge only when the charge reaches the negative terminal. But, even when the charge is in the middle of the circuit, why does the battery looses its "pumping" capability?

Answer 1

About the first question: The potential is created by the charges themselves and is found from their spatial distribution. It does not mean anything to say charges loose potential even though their energy can be deduced from the it. Moreover, The potential can be found by solving Poisson equation as illustrated in the following example:

Suppose 1000 electrons arrives at a bulb made of tungsten through a wire of copper. A fraction(say a 100) of them are going to interact with the bulb material to generate the light which accounts for a loss of electrical energy. This loss of energy is microscopically accounted for by the fact that the interacting electrons are going to be trapped in lower bands(cf. solid state physics). Consequently, these electrons will not participate to the flow of current leaving the bulb through another wire of copper. The balance sheet of this picture is that you have initially 1000 electrons arriving at the bulb and only 900 leaving it. The missing 100 electrons can be seen as 100 positive charges creating an electric field E computed from : $\mathbf{div}.\mathbf{E}=\frac{100e/V_{b}}{\epsilon_{0}} $ .

Where $V_b$ is the volume of the bulb. The spatial distribution of the potential U(r) can be then be deduced with: $\mathbf{E}=-\mathbf{grad}U $.

Eventually, the first 2 equations can be combined to get the Poisson equation : $\triangle U=\frac{100/V_{b}}{\epsilon_{0}} $

Solving it, we would end up with a drop of potential across the bulb related to the number of electrons lost during the lighting process. This drop of potential would have by the way a parabolic distribution if the 100-electrons distribution is assumed uniform.

Answer 2

Second question answered first: in context of physics or electrical engineering, a battery is modeled as something that provides a voltage difference, a difference in electrical potential, that can keep that voltage (nearly) no matter what you do to it.

So, some charge traveling (== some current flowing) will have no effect.

Real live batteries are different, and what you have probably in mind is a capacitor.

About the first question: some charge moving down the line of electrical potential (which need not to be a straight geometric line but will be bent depending on how the wires and resistors used are structured) will lose electrical potential energy. Anyway, what an electrical charge feels is not the electrical potential, but the directional potential difference.